# Prove that a certain integration yields the value $\frac{7}{9}$

Numerical methods surely indicate that $$\int_0^{\frac{1}{3}} 2 \sqrt{9 x+1} \sqrt{21 x-4 \sqrt{3} \sqrt{x (9 x+1)}+1} \left(4 \sqrt{3} \sqrt{x (9 x+1)}+1\right) \, dx= \frac{7}{9}$$.

Can this be formally demonstrated?

Yes it can, because the long square root equals $$\sqrt{9x+1}-\sqrt{12 x}$$. After this observation you get a standard integral, which reduces to an integral of rational function if you change the variable to $$\sqrt{9/x+1}$$.
Well, pursuing (as best I can) the line of reasoning put forth by K B Dave in his (stereographic-projection-motivated) answer to the related (antecedent) question in https://math.stackexchange.com/questions/3327016/can-knowledge-of-int-fx2-dx-possibly-be-used-in-obtaining-int-fx-dx, I make the transformation $$x\to \frac{4 m^2}{3 \left(m^2-3\right)^2}$$, having a jacobian of $$\frac{8 m \left(m^2+3\right)}{3 \left(m^2-3\right)^3}$$ and new limits of integration (0,1}, whereupon Mathematica directly yields the desired answer of $$\frac{7}{9}$$.
Let $$\begin{equation} 3 x=u^2, 9 x+1=v^2, \end{equation}$$ and stereographically project the curve $$v^2-3 u^2=1$$ about $$(u,v) =(0,1)$$, so that $$\begin{equation} \frac{v-1}{u}= \frac{3 u}{v+1}=m. \end{equation}$$
Then the integrands reduce to rational expression in $$m$$.,…